Question

integrate the function:
$\frac{1}{ \sqrt{ {(2 - x)}^{2} + 1} }$
​

Answer 1

Answer:

Explanation: It is one of the Standard Integral : ∫11+x2=arctanx+C .

Answer 2

Let 2-x=t

differentiating both sides w.r.t.x

$0 - 1 = \frac{dt}{dx}0−1=$

dx = - dtdx=−dt

Integrating the function w.r.t.x

$∫ \frac{1}{ \sqrt{ {(2 - x)}^{2} + t} }∫$

Put the value of 2-x=t and dx=-d

= $∫ \frac{ - dt}{ \sqrt{ {t}^{2} + 1} }=∫$

It is the form of :

$∫ \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } dx =$

$log |x + \sqrt{ {x}^{2} + {a}^{2} } |∫$

dx=log∣x+ x 2 +a2 ∣

∴ Replace a by 1 and x by t we get

= $- log |t + \sqrt{ {t}^{2} + 1 }$

= $log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1}$

= $log \frac{1}{ |t + \sqrt{ {t}^{2} + 1} | }$

= $log \frac{1}{ |2 - x + \sqrt{ {(2 - x)}^{2} + 1 } | }=$

= $log \frac{1}{ |2 - x + \sqrt{4 + {x}^{2} - 4x + 1} | }=log$

= $log \frac{1}{ |2 - x + \sqrt{ {x}^{2} - 4x + 5} | }$